Show $\mathfrak{gl}(n,\mathbb{R})_{\mathbb{C}}\cong \mathfrak{u}(n)_{\mathbb{C}}$

Question

I would appreciate help showing the complexification of these Lie algebras are isomorphic:

$$\mathfrak{gl}(n,\mathbb{R})_{\mathbb{C}}\cong \mathfrak{u}(n)_{\mathbb{C}}$$

where $\mathfrak{gl}(n,\mathbb{R})= M_n(\mathbb{R})$ and $\mathfrak{u}(n)=\lbrace X\in M(\mathbb{C}):X+X^{*}=0\rbrace$

Revealing my ignorance, I especially am confused regarding the complexification of $\mathfrak{u}(n)$ since it already is comprised of complex matrices. And how can $\mathfrak{gl}(n,\mathbb{R})_{\mathbb{C}}$ be isomorphic to $\mathfrak{u}(n)_{\mathbb{C}}$ in view of the stipulation regarding elements of $\mathfrak{u}(n) $that $X+X^{*}=0$

Thanks

Answer 1

Let $H_i$ denote the $n$-by-$n$ diagonal matrix with $1$ at the $(i,i)$-entry and $0$ elsewhere. For $i,j=1,2,\ldots,n$ with $i<j$, let $X_{i,j}$ be the matrix with $1$ at the $(i,j)$-entry and $0$ elsewhere, whilst $Y_{i,j}$ denotes $X_{i,j}^\top$. Note that the matrices $H_i$'s, $X_{i,j}$'s, and $Y_{i,j}$'s form a Chevalley basis of the Lie algebra $\mathfrak{gl}(n,\mathbb{C})$, noting that $$\mathfrak{gl}(n,\mathbb{R})_\mathbb{C}=\mathbb{C}\,\underset{\mathbb{R}}{\otimes}\,\mathfrak{gl}(n,\mathbb{R})\cong\mathfrak{gl}(n,\mathbb{C})\,.$$ (An isomorphism $\mathbb{C}\,\underset{\mathbb{R}}{\otimes}\mathfrak{gl}(n,\mathbb{R})\cong\mathfrak{gl}(n,\mathbb{C})$ should be easy to find).

A basis for $\mathfrak{u}(n)\subseteq\mathfrak{gl}(n,\mathbb{C})$ is given by $\tilde{h}_i$, $\tilde{x}_{i,j}$, and $\tilde{y}_{i,j}$ where $i,j=1,2,\ldots,n$ and $i<j$, where $$\tilde{h}_i:=\sqrt{-1}\,H_i\,,\,\,\tilde{x}_{i,j}:=\frac{1}{\sqrt{2}}\,\left(X_{i,j}-Y_{i,j}\right)\,,\text{ and }\tilde{y}_{i,j}:=\frac{\sqrt{-1}}{\sqrt{2}}\,\left(X_{i,j}+Y_{i,j}\right)\,.$$ Now, set $$h_i:=1\,\underset{\mathbb{R}}{\otimes}\,\tilde{h}_i\,,\,\,x_{i,j}:=1\,\underset{\mathbb{R}}{\otimes}\,\tilde{x}_{i,j}\,,\text{ and }y_{i,j}:=1\,\underset{\mathbb{R}}{\otimes}\,\tilde{y}_{i,j}\,.$$ Then, we see that the elements $h_i$'s, $x_{i,j}$'s, and $y_{i,j}$'s form a Chevalley basis of $\mathfrak{u}(n)_\mathbb{C}=\mathbb{C}\,\underset{\mathbb{R}}{\otimes}\,\mathfrak{u}(n)$. Consequently, an isomorphism $\varphi:\mathfrak{gl}(n,\mathbb{C})\to\mathfrak{u}(n)_\mathbb{C}$ is the linear extension of the assignments $$\varphi\left(H_i\right)=h_i\,,\,\,\varphi\left(X_{i,j}\right)=x_{i,j}\,,\text{ and }\varphi\left(Y_{i,j}\right)=y_{i,j}$$ for $i,j=1,2,\ldots,n$ with $i<j$.

P.S.: I said something wrong. I called the bases I provided Chevalley bases, but they are actually not.

Answer 2

This answer is incorrect - See comment below

In an attempt to answer my own question, can I not say that two finite dimensional vector spaces with the same dimension are isomorphic?

Thus $\dim \mathfrak{gl}(n,\mathbb{R})= \dim \mathfrak{u}(n)= n^2$ And their respective complexifications have dimension $2n^2$.